Question

If A =diag (2, -1,3), B=diag (-1,3, 2), then
AB =​

Answer 1

$\large\underline{\sf{Given- }}$

Two matrices A and B

$\red{\rm :\longmapsto\:A = diag \: [2 \: \: - 1 \: \: 3 \: ]}$

$\red{\rm :\longmapsto\:B = diag \: [ - 1 \: \: 3 \: \: 2 \: ]}$

$\green{\large\underline{\sf{To\:Find - }}}$

$\rm :\longmapsto\:AB$

$\blue{\large\underline{\sf{Solution-}}}$

Given that,

${\rm :\longmapsto\:A = diag \: [2 \: \: - 1 \: \: 3 \: ]}$

$\rm :\longmapsto\:A = \begin{gathered}\sf \left[\begin{array}{ccc}2&0&0\\0& - 1&0\\ 0&0&3\end{array}\right]\end{gathered}$

and

${\rm :\longmapsto\:B = diag \: [ - 1 \: \: 3 \: \: 2 \: ]}$

$\rm :\longmapsto\:B = \begin{gathered}\sf \left[\begin{array}{ccc} - 1&0&0\\0& 3&0\\ 0&0&2\end{array}\right]\end{gathered}$

Now, Consider

$\rm :\longmapsto\:AB$

$\rm \: = \: \begin{gathered}\sf \left[\begin{array}{ccc}2&0&0\\0& - 1&0\\ 0&0&3\end{array}\right]\end{gathered}\begin{gathered}\sf \left[\begin{array}{ccc} - 1&0&0\\0& 3&0\\ 0&0&2\end{array}\right]\end{gathered}$

$\rm \: = \: \begin{gathered}\sf \left[\begin{array}{ccc} - 2 + 0 + 0&0 + 0 + 0&0 + 0 + 0\\0 + 0 + 0& 0- 3 + 0&0 + 0 + 0\\ 0 + 0 + 0&0 + 0 + 0&0 + 0 + 6\end{array}\right]\end{gathered}$

$\rm \: = \: \begin{gathered}\sf \left[\begin{array}{ccc} - 2&0&0\\0& - 3&0\\ 0&0&6\end{array}\right]\end{gathered}$

$\rm \: = \: diag \: [ \: - 2 \: \: - 3 \: \: 6 \: ]$

Hence,

$\rm\rm \implies\: \:\boxed{ \tt{ \: AB = \: diag \: [ \: - 2 \: \: - 3 \: \: 6 \: ] \: }}$

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More to know :-

1. Matrix multiplication is defined when number of columns of pre multiplier is equal to number of rows of post multiplier.

2. Matrix multiplication may or may not be Commutative.

3. Matrix multiplication is Associative. i.e. A(BC) = (AB)C

4. Matrix multiplication is Distributive. i.e. A(B + C) = AB + AC